Problem: The lifespans of gorillas in a particular zoo are normally distributed. The average gorilla lives $23.6$ years; the standard deviation is $5.1$ years. Use the empirical rule (68-95-99.7%) to estimate the probability of a gorilla living less than $38.9$ years.
Explanation: $23.6$ $18.5$ $28.7$ $13.4$ $33.8$ $8.3$ $38.9$ $99.7\%$ $0.15\%$ $0.15\%$ We know the lifespans are normally distributed with an average lifespan of $23.6$ years. We know the standard deviation is $5.1$ years, so one standard deviation below the mean is $18.5$ years and one standard deviation above the mean is $28.7$ years. Two standard deviations below the mean is $13.4$ years and two standard deviations above the mean is $33.8$ years. Three standard deviations below the mean is $8.3$ years and three standard deviations above the mean is $38.9$ years. We are interested in the probability of a gorilla living less than $38.9$ years. The empirical rule (or the 68-95-99.7 rule) tells us that $99.7\%$ of the gorillas will have lifespans within 3 standard deviations of the average lifespan. The remaining $0.3\%$ of the gorillas will have lifespans that fall outside the shaded area. Because the normal distribution is symmetrical, half $({0.15\%})$ will live less than $8.3$ years and the other half $({0.15\%})$ will live longer than $38.9$ years. The probability of a particular gorilla living less than $38.9$ years is ${99.7\%} + {0.15\%}$, or $99.85\%$.